In first order reaction, the concentration of the reactant decreases from $$0.8$$ M to $$0.4$$ M in $$15$$ minutes. The time taken for the concentration to change from $$0.1$$ M to $$0.025$$ M is

First-Order Half-Life ($$t_{1/2}$$)

For a first-order reaction, the half-life ($$t_{1/2}$$) is a constant value. It depends solely on the rate constant ($$k$$) and is completely independent of the initial concentration of the reactant:

$$t_{1/2} = \frac{0.693}{k}$$

This means that no matter what the starting concentration is, it will always take the exact same amount of time for that concentration to decrease by half.

Step 1: Identify the half-life ($$t_{1/2}$$)

The problem states that the concentration decreases from $$0.8 \text{ M}$$ to $$0.4 \text{ M}$$ in $$15 \text{ minutes}$$.

Since $$0.4 \text{ M}$$ is exactly half of $$0.8 \text{ M}$$, this time interval represents one half-life:

$$t_{1/2} = 15 \text{ minutes}$$

• Step 2: Track the concentration decay from $$0.1 \text{ M}$$ to $$0.025 \text{ M}$$

  Let's follow the step-by-step half-life breakdown starting from an initial concentration of $$0.1 \text{ M}$$:

  • After the 1st half-life ($$15 \text{ min}$$): The concentration drops by half.

    $$0.1 \text{ M} \xrightarrow{t_{1/2}} 0.05 \text{ M}$$

  • After the 2nd half-life ($$s\text{an additional } 15 \text{ min}$$): The concentration drops by half again.

    $$0.05 \text{ M} \xrightarrow{t_{1/2}} 0.025 \text{ M}$$

The process requires exactly two consecutive half-lives ($$n = 2$$) to reach the target concentration:

$$\text{Total Time} = 2 \times t_{1/2}$$

$$\text{Total Time} = 2 \times 15 \text{ minutes} = 30 \text{ minutes}$$

Answer: Option A — 30 minutes